Math, asked by nandinibhardwaj19, 7 months ago

hey mates please solve these questions.
question 5 and 7 it's urgent please solve this.
please it's urgent

refrain from posting meaningless answer otherwise it will be reported ​

Attachments:

Answers

Answered by spiderman2019
4

Answer:

Step-by-step explanation:

Note: For question 5, since i donot have m in brainly superscript,  I am using x.

5) 49ⁿ⁺¹ * 7ⁿ - (343)ⁿ / 7³ˣ * 2⁴  = 3/343

=> (49)ⁿ * 49 * 7ⁿ -   (7³)ⁿ / 7³ˣ * 16 = 3 / 343

=> (7²)ⁿ * 7ⁿ * 49 -  (7³)ⁿ / 7³ˣ * 16 = 3 / 343

=> 7²ⁿ * 7ⁿ * 49 -  7³ⁿ / 7³ˣ * 16 = 3 / 343    

=> 7²ⁿ⁺ⁿ * 49 - 7³ⁿ/ 7³ˣ * 16 = 3/343                  

=> 7³ⁿ * 49 - 7³ⁿ / 7³ˣ * 16 = 3/343

//7³ⁿ can be taken as common in numerator on left hand side.

=> 7³ⁿ[ 49 - 1] / 7³ˣ . 16 = 3/343

=> 7³ⁿ * 48 / 7³ˣ . 16 = 3/343

//Since 16 * 3 = 48

=> 7³ⁿ * 3 / 7³ˣ = 3/343

//3 on both sides cancels out

=> 7³ⁿ⁻³ˣ = 1/343

=> 7³ⁿ⁻³ˣ = 7⁻³

//when bases are same powers are equal

=> 3(n - x ) = - 3

=> n - x = - 1

=> x = n + 1    ( Please replace m in place of x everywhere above)

7. (b³c⁻²/b⁺⁴c³)⁻³ ÷ (b⁻¹c/b²c⁻²)⁵ = bˣcᵃ   ( considering a instead of y)

Let us first solve L.H.S:

(b³ * b⁴ / c³ * c²)⁻³ ÷ (c * c²/ b² * b)⁵

=> [(b⁷/c⁵)⁻³] ÷  [(c³/b³)⁵]

=> [1 / (b⁷/c⁵)³] ÷  [(c³/b³)⁵]

=> [1 / (b²¹/c¹⁵)] ÷  [(c¹⁵/b¹⁵)]  

=> [c¹⁵/b²¹]   ÷ [(c¹⁵/b¹⁵)]

=> [c¹⁵ / b²¹]  * [b¹⁵/c¹⁵]

=> c¹⁵⁻¹⁵ / b²¹⁻¹⁵

=> c⁰/b⁶

=> b⁻⁶c⁰

Now,

b⁻⁶c⁰ = bˣcᵃ                           ( Please substitute a with y)

=> x = - 6, a = 0

x + a + 6 = - 6 + 0 + 6 = 0  

Hence proved.  

Similar questions